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A right circular cone is inscribed in a hemisphere so that the base of

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Bunuel
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A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   Updated on: 01 Jun 2021, 12:56
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A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)
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B

Originally posted by Bunuel on 03 Apr 2012, 16:49.
Last edited by Bunuel on 01 Jun 2021, 12:56, edited 3 times in total.
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   09 Jul 2012, 04:41
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SOLUTION

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.


Answer: B.

Show Spoiler
Attachment:
Cone.png
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   09 Jul 2012, 04:42
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Check other 3-D Geometry Questions from our Special Questions Directory.
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   23 Nov 2012, 02:10
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Not sure if anyone else was scratching their heads wondering why the height couldn't be less than the radius, but just in case... it has to do with the word "inscribed": to draw within a figure so as to touch in as many places as possible <a regular polygon inscribed in a circle>

lol.
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   08 Jun 2013, 22:36
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It's the combination of "right circular cone" and "inscribed". The cone shares a base with the hemisphere and it's tip touches the top of the hemisphere. If you had a right circular cone of a height to radius ratio of anything other than 1:1 you would not be able to inscribe it in a hemisphere. Essentially, what you've described is impossible for this particular problem.

Also, you cannot have a hemisphere of the type you've described. It wouldn't be a hemisphere. A sphere's radius is the same throughout. A hemisphere is half of a sphere (so it's height must equal it's radius.
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   06 May 2014, 12:57
Bunuel wrote:
SOLUTION

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.
Attachment:
Cone.png


Answer: B.

Hi Bunuel, how did you conclude the underlined in above statement? Thanks!
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   07 May 2014, 04:25
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Dienekes wrote:
Bunuel wrote:
SOLUTION

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.
Attachment:
The attachment Cone.png is no longer available


Answer: B.

Hi Bunuel, how did you conclude the underlined in above statement? Thanks!


Consider the cross-section. We'd have an isosceles triangle inscribed in a semi-circle:
Attachment:
Untitled.png
Untitled.png [ 2.44 KiB | Viewed 66880 times ]
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   19 Jul 2016, 11:30
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height and radius are equal ...follow fig.

ans B
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   20 Apr 2017, 19:04
Bunuel wrote:
Dienekes wrote:
Bunuel wrote:
SOLUTION

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.
Attachment:
The attachment Cone.png is no longer available


Answer: B.

Hi Bunuel, how did you conclude the underlined in above statement? Thanks!


Consider the cross-section. We'd have an isosceles triangle inscribed in a semi-circle:
Attachment:
The attachment Untitled.png is no longer available


Why cone can't be like the image below?
Attachment:
File comment: cone in hemisphere
cone in hemisphere.png
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   20 Apr 2017, 21:28
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avinashvpec wrote:
Why cone can't be like the image below?
Attachment:
cone in hemisphere.png


We are told that the cone is "inscribed" in a hemisphere, not just inside, so the vertex must be on the circumference.
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   07 Jul 2017, 06:06
Bunuel

It makes sense that the height of the cone will be equal to the radius of the hemisphere.

However, when I tried equating the volumes for both the cone and hemisphere, the solution I am getting is 2 : 1

1\3 x pie x r^2 x h = 1/2 x 4/3 x pie x r^3

h = 2r


Am I missing something here. Please help
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   07 Jul 2017, 06:23
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kotharishakti wrote:
Bunuel

It makes sense that the height of the cone will be equal to the radius of the hemisphere.

However, when I tried equating the volumes for both the cone and hemisphere, the solution I am getting is 2 : 1

1\3 x pie x r^2 x h = 1/2 x 4/3 x pie x r^3

h = 2r


Am I missing something here. Please help


Why are you equating volumes? The volume of a cone is obviously less than the volume of the hemisphere it is inscribed in.
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   07 Jul 2017, 07:20
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Bunuel wrote:
kotharishakti wrote:
Bunuel

It makes sense that the height of the cone will be equal to the radius of the hemisphere.

However, when I tried equating the volumes for both the cone and hemisphere, the solution I am getting is 2 : 1

1\3 x pie x r^2 x h = 1/2 x 4/3 x pie x r^3

h = 2r


Am I missing something here. Please help


Why are you equating volumes? The volume of a cone is obviously less than the volume of the hemisphere it is inscribed in.




Oh yes!!
That's Correct Bunuel.

I am doing it wrong! Such a silly one.

Thanks a lot for your reply!
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   27 Apr 2018, 10:03
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NvrEvrGvUp wrote:
A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

A. \(\sqrt{3} : 1\)

B. \(1 : 1\)

C. \(\frac{1}{2} : 1\)

D. \(\sqrt{2} : 1\)

E. \(2 : 1\)


If a right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere, then the height of the cone is exactly the radius of the hemisphere. So the ratio is 1:1.

Answer: B
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   13 Sep 2019, 14:49
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Bunuel wrote:
avinashvpec wrote:
Why cone can't be like the image below?
Attachment:
cone in hemisphere.png


We are told that the cone is "inscribed" in a hemisphere, not just inside, so the vertex must be on the circumference.



This is what stumped me. I thought the base of the cone needed to only touch the base of the sphere, not extend to the edges of the sphere. So in my situation my height was the radius.

If we are told that the bases "coincide" do we presume they align equally?
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   16 Dec 2019, 18:20
Bunuel wrote:
SOLUTION

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.


Answer: B.

Show Spoiler
Attachment:
Cone.png


Bunuel

Out of curiosity -->

Is the vertex angle (the one at the very top) -- can i assume that angle is a 90 degree angle ?

Per my understanding I can because the triangle is inscribed and the vertex angle thus HAS to be 90 degrees given the base of the triangle is the diameter of the sphere

Please let me know if my understanding is accurate ?

Thank you !
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   31 Mar 2020, 08:23
Experts - chetan2u, Bunuel, generis, MahmoudFawzy

Just wondering if you could reply to the above post right above this one ?
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   22 Apr 2020, 15:07
Bunuel wrote:
SOLUTION

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.


Answer: B.

Show Spoiler
Attachment:
The attachment Cone.png is no longer available


Hi Bunuel

How can one be sure the right angled cone isn't like this per the picture i have drawn ?

Angle BAC is 90 degree per my picture

If any issues with the picture, you should be able to zoom in

Thank you!
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   22 Apr 2020, 23:57
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jabhatta@umail.iu.edu wrote:
Bunuel wrote:
SOLUTION

A right circular cone is inscribed in a hemisphere so that the base of the cone coincides with the base of the hemisphere. What is the ratio of the height of the cone to the radius of the hemisphere?

(A) \(\sqrt{3}:1\)

(B) \(1:1\)

(C) \(\frac{1}{2}:1\)

(D) \(\sqrt{2}:1\)

(E) \(2:1\)

Note that a hemisphere is just a half of a sphere.

Now, since the cone is a right circular one, then the vertex of the cone must touch the surface of the hemisphere directly above the center of the base (as shown in the diagram below), which makes the height of the cone also the radius of the hemisphere.


Answer: B.

Show Spoiler
Attachment:
Cone.png


Hi Bunuel

How can one be sure the right angled cone isn't like this per the picture i have drawn ?

Angle BAC is 90 degree per my picture

If any issues with the picture, you should be able to zoom in

Thank you!


That's not a right circular cone. A right circular cone is the one where the vertex is above the base center point..
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Re: A right circular cone is inscribed in a hemisphere so that the base of [#permalink] New post   05 Feb 2022, 07:45
Having present hemisphere is half of a sphere.

If we inscribe a circular cone in a hemisphere, the base of the cone must coincide with the base of the hemisphere, both are identical circumferences.

Thus the radius of the base of the circular cone is the same radius as the base of the hemisphere.

Since the hemisphere is half that of a sphere, the height of the sphere coincides with the height of the cone.

The height of the sphere is the radius of the base of the hemisphere.

Therefore, the radius of the sphere is equal to the height of the cone with a circular base.

Answer B
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